Let f : left(-frac{pi}{2}, frac{pi}{2}right) | vedclass

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(A, C) Step $1$: Check for odd/even function.$f(-x) = (\log(\sec(-x) + 	an(-x)))^3 = (\log(\sec x - 	an x))^3$.Since $\sec x - 	an x = \frac{1}{\se

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$f(x)$ is a non-one-one function

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(A, C) Step $1$: Check for odd/even function.$f(-x) = (\log(\sec(-x) + 	an(-x)))^3 = (\log(\sec x - 	an x))^3$.Since $\sec x - 	an x = \frac{1}{\sec x + 	an x}$,we have $\log(\sec x - 	an x) = \log((\sec x + 	an x)^{-1}) = -\log(\sec x + 	an x)$.Thus,$f(-x) = (-\log(\sec x + 	an x))^3 = -(\log(\sec x + 	an x))^3 = -f(x)$.Therefore,$f(x)$ is an odd function.Step $2$: Check for one-one function.$f'(x) = 3(\log(\sec x + 	an x))^2 \cdot \frac{1}{\sec x + 	an x} \cdot (\sec x 	an x + \sec^2 x) = 3(\log(\sec x + 	an x))^2 \cdot \sec x$.Since $\sec x > 0$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$ and $(\log(\sec x + 	an x))^2 \ge 0$,$f'(x) \ge 0$. The function is strictly increasing,so it is a one-one function.Step $3$: Check for onto function.As $x ightarrow \frac{\pi}{2}^-$,$\sec x + 	an x ightarrow \infty$,so $f(x) ightarrow \infty$.As $x ightarrow -\frac{\pi}{2}^+$,$\sec x + 	an x ightarrow 0^+$,so $\log(\sec x + 	an x) ightarrow -\infty$,and $f(x) ightarrow -\infty$.Since the range is $(-\infty, \infty) = \mathbb{R}$,the function is onto.Conclusion: $f(x)$ is an odd function and an onto function.

Let $f : \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow \mathbb{R}$ be defined by $f(x) = (\log(\sec x + \tan x))^3$. Then:

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